Another radiation question

If a charge is accelerating, but you are also accelerating WITH the charge, will you see any radiation? I.e. will the integral of the Poynting vector over a closed surface surrounding the charge be zero? It seems that it should be.

Now lets say YOU are accelerating, but the charge is NOT. You woudn't see radiation from the charge in this case, will you?

for the first question, yes, you will see radiation
for the second question, No, you won't wee radiation

remember you are not in inertial reference fram in both case, you have no reason to expect the answer is same as you sitting in an inertial reference.... velocity is relative, but accelerating is absolute, which is independent to the observer's motion... the charge feels acceleration. and an accelerating charge emits radiation, period

it is a bit weird in the first case because you do not see the charge moving and it radiates, but again, you are not in the inertial frame, same as you see stuffs behave weird when were in an accelerating bus......

for the first question, yes, you will see radiation
for the second question, No, you won't wee radiation

remember you are not in inertial reference fram in both case, you have no reason to expect the answer is same as you sitting in an inertial reference.... velocity is relative, but accelerating is absolute, which is independent to the observer's motion... the charge feels acceleration. and an accelerating charge emits radiation, period

it is a bit weird in the first case because you do not see the charge moving and it radiates, but again, you are not in the inertial frame, same as you see stuffs behave weird when were in an accelerating bus......

Yes, yes, that's what I would have thought too (and you may very well be right). But lets say we have a charge sitting on a table in the earth's gravitational field. I don't think you'll measure any radiation, otherwise you'd have a source of "free" energy. Locally, this is equivalent to the charge being in an accelerating reference frame and you accelerating with the charge. So, I would assume that in the latter case the radiation you'd measure would also be zero.

So, I would assume that in the latter case the radiation you'd measure would also be zero.

latter case? did u mean the former case?
if you have a charge on earth and you are rotating with it... you WILL see radiation... just the rotation of the earth is too slow (1 rotation / day) that the radiation is far too small for us to detect it.....
and NO, you won't get free energy... the rotation of earth will slow down by the charge when it radiate, you have a price to pay.... the day is longer (by a tiny amount). again, the effect will be too small for us to detect.....

latter case? did u mean the former case?
if you have a charge on earth and you are rotating with it... you WILL see radiation... just the rotation of the earth is too slow (1 rotation / day) that the radiation is far too small for us to detect it.....
and NO, you won't get free energy... the rotation of earth will slow down by the charge when it radiate, you have a price to pay.... the day is longer (by a tiny amount). again, the effect will be too small for us to detect.....

I wasn't talking about rotation. I was talking about the equivalence principle which states that a gravitational field is locally equivalent to an accelerating reference frame.

For example Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969

Abstract: The connection between an accelerated charge an and one at rest in a (weak) gravitational field is discussed in accordance with the principle of equivalence principle. For that purpose, the fields produced by a freely falling charge and a supported one (i.e. at rest in a gravitational field) are transformed to the rest frame of the observer, who may be similarly supported or freely falling. A nonvanishing energy flux is found only if the charge is freely falling and the observer supported, or vice versa. This agrees with previously established results.

pmb_phy:
your article seems confirms what i said..... and it explain very clearly .....
did your conclustion based on that article? or it is your own thought.... If it is your own thought, why bring up something contradict your arguement

pmb_phy:
your article seems confirms what i said..... and it explain very clearly .....
did your conclustion based on that article? or it is your own thought.... If it is your own thought, why bring up something contradict your arguement

Please clarify. The question was

If a charge is accelerating, but you are also accelerating WITH the charge, will you see any radiation?

I responded "No." As the article states in the abstract

A nonvanishing energy flux is found only if the charge is freely falling and the observer supported or vise versa.

You didn't explain how what I said contradicted the articles I quoted/referanced.

I've already read the article the you referred to. The author contradcits a lot of physics in that article. For example; the author assumes, using his own intuition or whatever, that some forms of the equivlance principle don't apply. Since the equivalence principle is a postulate then only experimentation can determine whether it is correct or not - not calculation.

The author asserts

Since the particle is radiating energy which can be detected and used, conservation of energy suggests that the radiated energy must be furnished by the rocket — we must burn more fuel to produce a given accelerating worldline than we would to produce the same worldline for a neutral particle of the same mass.

This contradicts previous conclusions. For example; An extra force must be used to accelerate a charged particle over a non-charged particle. This extra force is known in the literature as the "self-force" acting on the particle. This also is known as the "radiation reaction force."

Details - The power radiated by an accelerating charge is given (in esu units) by P = (2/3)q2/c[sup3[/sup](a)2. To account for this energy loss modify Newton's equation by adding in an extra force Frad. Suppose the force on a non-charged particle of the same (rest) mass is (in the non-relativistic limit) Fext = ma. A particle with charge q and mass m is then given by Fext + Frad = ma. By demanding that energy be conserved this demands Frad = mTda /dt (where T is a constant and I'm too lazy to type it in :tongue: ). If the particle is uniformly accelerated then the radiative force is zero. Thus if you have a particle in a uniformly accelerating frame of reference and you're also at rest in that frame then the weight of the particle is independant of the charge. Hence the authors assertions are in contradiction with the assertions I've just stated.

Its also be shown that the weight of a particle depends on the spacetime curvature. That means that a charged particle at rest on, say, the earth will have a different weight (i.e. smaller required support force) than the same particle at rest in a uniform g-field.

I.e. will the integral of the Poynting vector over a closed surface surrounding the charge be zero? It seems that it should be.

pmb_phy said:

The Poyting vector will be zero.

The article Vinchentan cited says:

Although each observer in a rigidly accelerating elevator surrounding the particle measures a vanishing Poynting vector in his own private rest frame, nevertheless, taken as a whole there is radiation through the elevator walls. Adding the (zero) energy fluxes measured by each observer on the wall in his private rest frame to (incorrectly) conclude zero total energy radiation is an illegitimate operation because these energy fluxes refer to different rest frames.